Problem: Find one value of $x$ that is a solution to the equation: $(x^2-8)^2+x^2-8=20$ $x=$
Answer: We could solve for $x$ by expanding $(x^2-8)^2$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form. Let's look at the given equation: $({x^2-8})^2+1({x^2-8})=20$ If we let ${p}={x^2-8}$, we can see that this equation is in the form: ${p}^2+1{p}=20$ Let's solve this equation in terms of ${p}$ : $\begin{aligned}{p}^2+1{p}&=20\\\\ {p}^2+1{p}-20&=0\\\\ ({p}-4)({p}+5)&=0\\\\ {p}=4\ &\text{or} \ \ {p}=-5 \end{aligned}$ Since ${p}={x^2-8}$, let's substitute this value back into our two solutions in order to solve for $x$ : ${x^2-8}=4\ \ \ \text{or} \ \ \ {x^2-8}=-5$ When we solve ${x^2-8}=4$, we find that $x=\pm\sqrt{12}$. When we solve ${x^2-8}=-5$, we find that $x=\pm\sqrt{3}$. In conclusion, the four solutions of the equation $(x^2-8)^2+x^2-8=20$ are: $x=\sqrt{12}$ $x=-\sqrt{12}$ $x=\sqrt{3}$ $x=-\sqrt{3}$ [Is there another way to solve for x?]